Solving a Rational Equation with Monomial Denominators

Solving a Rational Equation with Monomial Denominators


Hi. I’ll walk you through this
example to explore solving a rational equation. Solving a rational equation
involves the following steps. Firstly, find all
restrictions. That means we will find all
the values that make any denominator in the equation 0. These values will be
avoided from being considered as solutions. So let us look at
the equation. The denominators in the given
equation are x and 2x. Look at the numbers negative
3 and positive 3. They can otherwise be written
as negative 3 over 1 and 3 over 1. The denominator for these
numbers then would be 1 in both cases. When you see 1 as
the denominator, you can ignore it. Now we look for values that
make the denominators 0. The denominators x and 2x will
become 0 when x is equal to 0. So the restriction will
be x is equal to 0. The second step is to find the
LCD of all the rational expressions in the
given equation. In order to do that, we look at
these denominators, x and 2x, and determine their LCD. Between x and 2x, we can
quickly recognize that the LCD is 2x. The third step is to multiply
the equation by LCD. The equation we have is 5/x
minus 3 equals 7/2x plus 3. We multiply this equation
by the LCD we found. So let us wrap the left-hand
side as well as the right-hand side, and multiply by the
LCD, which is 2x. We distribute the 2x on to
every term inside the parentheses on both sides, and
that would give us 2x times 5/x minus 2x times 3 equals 2x
times 7/2x plus 2x times 3. Now we will simplify each
term of this equation. In the first term of the
left-hand side, we see x on the top and x in the bottom. They divide each other,
giving 1. Same way in the first term of
the right-hand side, we see 2x on the top and 2x
in the bottom. They, too, divide each
other, giving 1. Now we simplify the equation
that we have, and that would be 2 times 5, which is 10, minus
2x times 3, and that will be 6x equals– we just have 7 in the first term
of the right-hand side– plus the second term
will give 6x For the next step, we take that
resulting equation and solve for x. You must observe that we got the
denominator as 1 for every term in that equation. We will now go ahead
and solve this. In order to do that, we group
x terms to one side and numbers to the other. Let us add 6x to both sides so
we can group the x terms. We are left with 10 on the
left-hand side, and 7 plus 12x on the right-hand side. Now to bring the numbers to the
same side, we subtract 7 from both sides. We will have 3 equals 12x. To isolate x, we divide by 12
because 12x is 12 times x, and we do the opposite to
get rid of that 12. So we divide by 12, and what we
do to one side we do to the other as well. And this would give
x to be 1/4. As step 5, we will compare the
proposed solutions to the list of restrictions and reject any
proposed solution that is in the list of restrictions. The solution we got
is x equals 1/4. And in the list of restrictions,
if you try to look at step 1, we have
restriction to be just 0. Since 1/4 is not equal to 0, we
do not reject the solution. As the last step, we go
ahead and check the validity of the solution. In order to do that, we plug
in the solution in the original equation and
look for truth. We got x equals 1/4. That’s the proposed solution. Let us plug that in the
original equation. The original equation is 5/x
minus 3 equals 7/2x plus 3. We are plugging 1/4
in the place of x. And this is what we get. We take that and try to
simplify it further. You must recall how to divide
when you have fractions in the denominator. So we make the division as
multiplication and flip the following fraction, which will
give us 5 times 4/1 minus 3 equals 7 times 4 divided
by 2 plus 3. Simplifying further, we will get
17 equals 17, which is a true statement. What that tells us is x equals
1/4 is a valid solution for the given equation. I hope this helps. Thank you.

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